Chapter 12 12.pdf · 12.7 Elasticity: Tension and Compression • For simple tension or...
Transcript of Chapter 12 12.pdf · 12.7 Elasticity: Tension and Compression • For simple tension or...
Chapter 12
Equilibrium and Elasticity
HyperPhysics: Equilibrium Simple Machines
12.2 Equilibrium
• What allows objects to be stable in spite
of forces acting on it?
• Under what conditions do objects
deform?
The two requirements for the state of
equilibrium are:
1. The linear momentum of the center of
mass is constant.
2. The angular momentum about the center of
mass, or about any other point, is also
constant.
The balancing rock of Fig. 12-1 is an example of an
object that is in static equilibrium. That is, in this
situation the constants in the above requirements
are zero.
Fig. 12-1 A balancing rock. Although its perch
seems precarious, the rock is in static
equilibrium. (Symon Lobsang/Photis/ Jupiter
Images Corp.)
The ladder and the forces on it: Forces in both directions sum to 0:
The torques are all perpendicular
to the plane of the page, so there’s
only one torque equation:
Solve the 3 equations to get:
Force, x: n1 n2 0
Force, y: n1 mg 0
Torque: Ln2 sin L
2mgcos 0
tan sin
cos
1
2
12.4 Example: Static equilibrium, leaning ladder
• At what angle will the ladder slip?
Clicker question The figure shows a person in static equilibrium leaning against
a wall. Which one of the following must be true?
A. There must be a frictional force
at the wall, but not necessarily
at the floor.
B. There must be a frictional force
at the floor, but not necessarily
at the wall.
C. There must be frictional forces
at both wall and floor.
12.7: Elasticity
• A stress is defined as deforming force per unit area, which produces a
strain, or unit deformation.
• Stress and strain are proportional to each other. The constant of
proportionality is called a modulus of elasticity.
Stress = Modulus x Strain
12.7 Elasticity: Tension and Compression
• For simple tension or compression, the
stress on the object is defined as F/A,
where F is the magnitude of the force
applied perpendicularly to an area A on
the object.
• The strain, or unit deformation, is then the
dimensionless quantity ΔL/L, the fractional
change in a length of the specimen.
• The modulus for tensile and compressive
stresses is called the Young’s modulus
and is represented in engineering practice
by the symbol E.
Fig. 12-12 A stress–strain curve for a steel test
specimen. The specimen deforms permanently
when the stress is equal to the yield strength of the
specimen’s material. It ruptures when the stress is
equal to the ultimate strength of the material.
12.7: Elasticity: Shearing
• In the case of shearing, the stress is also a force per unit area, but the force
vector lies in the plane of the area rather than perpendicular to it.
• The strain is the dimensionless ratio Δx/L, with the quantities defined as
shown in the figure.
• The corresponding modulus, which is given the symbol G in engineering
practice, is called the shear modulus.
12.7 Elasticity: Hydraulic Stress
• In the figure, the stress is the fluid pressure p on the object, where
pressure is a force per unit area.
• The strain is ΔV/V, where V is the original volume of the specimen
and ΔV is the absolute value of the change in volume.
• The corresponding modulus, with symbol B, is called the bulk
modulus of the material. The object is said to be under hydraulic
compression, and the pressure can be called the hydraulic stress.
Example: wobbly table
We take the table plus steel cylinder as our system.
The situation is like that in the figure. If the tabletop
remains level, the legs must be compressed in the
following ways: Each of the short legs must be
compressed by the same amount (call it ΔL3) and thus
by the same force of magnitude F3.The single long leg
must be compressed by a larger amount ΔL4 and thus
by a force with a larger magnitude F4.